1) Titration reveals that 11.6 mL of 3.0 M sulfuric acid are required to neutral

ID: 1011070 • Letter: 1

Question

1) Titration reveals that 11.6 mL of 3.0 M sulfuric acid are required to neutralize the

sodium hydroxide in 25.00 mL of NaOH solution. What is the molarity of the NaOH

solution?

Please show work:) Also I know this might be too much but can you please answer question 2 as well! Thanks

2)When 34.2 mL of a 1.02 M NaOH solution is added from a buret to 25.00 mL of a

phosphoric acid solution that contains phenolphthalein, the solution changes from

colorless to red. What is the molarity of the phosphoric acid?

Moles of sulfuric aicd in 3 M and 11.6 ml = molarity* volume in L = 3*11.6/1000=0.0348 moles

the reaction between NaOH and H2SO4 is

2NaOH + H2SO4---------> Na2SO4 +2H2O

moles of NaOH= 1/2 times moles of H2SO4= 1/2*0.0348=0.0174

molairt of NaOH= 0.0174*1000/25 =0.696 M

Moles of NaOH= Molarity* Volume (L)= 1.02*34.2/1000 =0.035

The reaction between H3PO4 and NaOH can be represented as

H3PO4 +3NaOH-----------> Na3PO4+3H2O

3 mole of NaOH requires 1 mole of H3PO4

0.035 mole of NaOH requires 0.035/3=0.0117

Molarity of H3PO4= 0.0117*1000/25=0.467

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